Sunday, March 28, 2010

I said in my last post that I was going to generalize the helium atom so I could calculate the ground state energy for the whole isoelectonic series. The first step is to work out the helium atom.

The key to working out the energy is pretty simple (in retrospect! it was hard for me to do it the first time). There are three pieces: the kinetic energy, the potential energy, and the interaction energy (which comes from the electron-electron repulsion.) The starting point will be what I called the "naive model", where we take the wave functions to be hydrogen like, we drop both electrons into the ground state, and add the interaction energy as an afterthought. Here is how it goes.

With just one electron, helium scales from the hydrogen atom by a factor of two on linear dimension, which gives a factor of four on energy. So where hydrogen is -1 Ry, He+ (singly ionized) is -4 Ry. This is true theoretically and experimentally; it's pretty much an exact result.

We can look inside this quantity of energy and break it down further into kinetic and potential. This is the same thing you can do with a planet orbiting the sun. The kinetic energy is positive and the potential energy is negative. There is something called the virial theorem which tells you to expect these quantities to be in the ration of 2:1; it works for planets, and without getting into the reasons why, suffice to say it works for atoms too. The kinetic energy of He+ is +4 Ry and the potential is -8 Ry. Total: -4 Ry.

Now we add the second electron and assume it goes into exactly the same energy state as the first. Total energy of the atom is now -8 Ry. But that's neglecting the mutual repulsion of the electrons. As I mentioned in my last post, that's an exact integral that comes to 2.5 Ry, giving a grand total of -5.5 Ry. Let's write it out:

That's what I've called the "naive model". Now we're ready to have some fun.

The first thing to do is to optimize the wave function to try and minimize the total energy. The calculation above assumes that the wave function is just the same as the hydrogen wave function, scaled by a factor of two on linear dimension. In fact, because the two electrons are crowding each other, we might expect the wave function to spread out a bit to try and compensate. The wave function is essentially of the form exp(-r), and the easiest way to spread it out is to put a variable parameter k into the exponential decay. The nice thing is that with this simple change, the three terms in our energy equation are altered in a very simple way.

(The traditional explanation is a little different, as you can see on the Wikipedia page. They talk about mutual screening and "effective nuclear charge Z". This is confusing for a couple of reasons, mostly because Z is normally reserved for nuclear charge number and later I'm going to need to vary just that parameter. So I stick with what I'm calling a relaxation parameter, k, which just relaxes the shape of the wave function without any particular justification.)

The question is, how does my relaxation factor change the energy? Looking at it term by term, it's pretty simple. The kinetic energy is quadratic in k and the other terms are linear. Why? The kinetic is quadratic because in quantum mechanics you differentiate twice to get KE, and each differentiation gives a factor of k. The potential is linear because of the 1/r term in the equation for potential energy. Change the average distance by a factor of k, and the energy goes by the same factor. So we can write, without any particular difficulty, the revised energy equation as follows:

You have got to check out Wikipedia to see how much work they went through to get from (eqn. 1) to (eqn. 2). I did it with virtually no work at all by using a simple scaling argument. Once you're here, it's a very simple first-year calculus exercise to idenfiy the minium energy as occuring when k= 27/32. (On Wikipedia they get 27/16, but they define their optimization parameter a little differently.) You get Energy = 729/128 Ry, or just about 77.5 eV. It's about 1.5 eV short of the true energy.

This is where they basically leave off, letting you think that if you keep tinkering with the shape of the wave function, you'll get closer and closer to the true energy. You won't. You have to do something really radical to improve on this calculation much beyond the point where we've already taken it. You have to come up with a wave function that includes the coordinates of both electrons independently. But that's a story for another day.

What we're going to do next is take the optimized function for helium and generalize it to the rest of the isoelectronic series. We're going to need to introduce a parameter Z for nuclear charge. And we're going to think about how the energy terms in (eqn. 2) scale with Z.

It's going to be pretty easy, but it's just a little different than when we scaled to k. This time, the potential and kinetic terms are both going to be quadratic, and only the interaction will be linear. It works this way because a factor of Z on nuclear charge gives you a factor of 1/Z on linear dimension. For kinetic energy, this gives you Z-squared because of the double differentiation. For potential energy, you get a factor of Z for dimension and another factor of Z for the increased charge: hence, Z-squared. For the interaction however, although you get a factor of Z on dimension, the charges are the same two electrons you started out with, so there's no additional factor. It's just linear in Z.

So we take eqn. 2 and just put Z or Z-squared in front of the terms as appropriate:

Saturday, March 27, 2010

I ended up writing a paper about what I figured out for helium. The paper starts off following pretty much the analysis of the helium atom that you'll find in the Wikipedia article. (There's actually a better version of this analysis on a University of Texas website http://farside.ph.utexas.edu/teaching/qmech/lectures/node128.htmlbut I'm just going to refer to Wikipedia for convenience.) There are basically three stages in the refinement of the analysis:

1. First case, you treat it as a hydrogen atom with two electrons in the same ground orbital. Scaling for the charge and reduced dimension, you get a ground state energy or -108.8 eV, or exactly -8 Rydbergs (where hydrogen is -1 Ry).

2. First improvement: you keep the same wave function as in case (1), but you calculate the repulsion energy of the two electrons. It turns out to be an exact integral, and it comes to 2.5 Ry. (It's not hard to do numberically either because of the spherical symmetry.) So the refined estimate of the energy is -5.5 Ry, or -74.8 eV. It's actually pretty close to the experimental value of -79.0 eV. I call this the "naive model".

3. Second improvement. You tweak the wave function by putting a variable parameter in front of the exponential decay. The nice thing is because you're still basically using the same wave function except for a scaling factor, you don't have to redo any of the hard 3-dimensional integrals. You just scale the ones you already have.

When you do case (3), you get an optimization parameter that lets you bring the energy down to -77.5 eV. Now you're within 1.5 eV of the true value. The funny thing is, why didn't you get even closer? The "naive model" was already within 4 eV of the true value, so this optimization trick has really only closed the gap by 60%.

That's why I find it odd what the guy at U of Texas says at the end of his article. He says: "Obviously, we could get even closer to the correct value of the helium ground-state energy by using a more complicated trial wave-function with more adjustable parameters. ". It's a funny statement because how obvious is it? Based on the numbers so far it doesn't appear to me that we're closing in fast enough.

In hindsight, I know the reason. You can make the wave function as complicated as you want and it will only get you so close and no closer. To get the correct ground state energy you have to do something quite tricky and unexpected, at least based on the analysis so far I'd have to say it's unexpected. But I'll get to that a little later.

What I did in my analysis was to generalize the equations in the Wikipedia article so they work on any size nucleus, as long as it has two electrons. So it works for the hydrogen negative ion (H-); it works for neutral helium; it works for singly ionized Lithium (Li+); and it works for doubly ionized beryllium (Be++). It works for all the atoms of course, but the first four are the ones I was able to look up and check the actual energies.

But before we get to my results, I'm going to brag about how clever I was in deriving these equations. I did a bunch of stuff using basic scaling principles that let me go from A to B in a single line when Wikipedia (and Texas) use a whole page of hard math to do the same thing. I think I'll tell you about that in my next post.

Friday, March 26, 2010

In my last post, I talked about how the solution of the two-electron well becomes simple in the two extreme cases: the very large box, and the very small box. And how these cases correspond to having a fixed size of box while varying the strength of the electric repulsion. The funny thing is, the very small box corresponds to the case of weak repulsion, and the very big box is the case of strong repulsion. Why is this?

The answer turns out to be almost obvious: The repulsive energy is a 1/r force, but the kinetic energy goes as 1/r-squared. You know about the 1/r from basic electrostatics; you see it in the fomula for the potential of a spherical charge. It comes from the 1/r-squared law of forces. Since energy is the integral of force along a path, it always comes to something in 1/r for spherical geometry. The kinetic energy, however, is a purely quantum-mechanical thing. For the particle confined in a box, the kinetic energy is the square of the momentum operator. The momentum operator is differention (well, it's the del operator in three dimension). If you make the box twice as small, the derivatives are twice as big, so when applied twice, you get a factor of 4. In other words, it's an inverse square relationship on the dimension of the containment. So for a very small box the kinetic term dominates the potential term.

The interesting thing is that you can see this in a physical case: it's called the isoelectronic series of helium, and it goes all the way back to my second post about why the helium atom doesn't have a miniature replica in the hydrogen atom. It has to do with the scaling effect being different for the potential and kinetic energy terms, and I did some really cool calculations to show how this works for certain atomic energy levels: hence, "the isoelectronic series of helium".

You may know that when you solve the equation for the hydrogen atom, you automatically get by analogy the solution for a whole series of other atoms in their ionized states, when they have only one electron. For example, singly ionized helium is exactly the same as hydrogen except the wave functions are compressed by a factor of two and all the energy levels are four times greater; ditto for doubly-ionized lithium, except it's three times on size and nine times on energy. Etcetera. It's called the isoelectronic series of hydrogen.

The series for helium is a little different. You don't get the same geometric scaling effect, and the reason is because you're basically putting two eletrons in a box, and it makes a difference how big the box is. You can see it pretty clearly if you look up (as I did) the energy levels for the atomic states belonging to this sequence.

Wednesday, March 24, 2010

When I left off yesterday, I had sketched out the proposed symmetric solution of the two-electron potential well. I got it by taking two simple solutions where the electrons crowded themselves to opposite sides of the box, and adding those two solutions together.

The peculiar thing is there is another possible solution that you have to also consider: it's called the anti-symmetric solution, and you get it by taking the difference instead of the sum. The wave function sketched in the coordinates of the two electrons looks like this:And the question is: which one has lower energy - the symmetric or the antisymmetric combination? And it's not totally obvious to me what the answer should be.

So I posted this question on an interned forum, and one guy gave the very helpful suggestion of "turning off" the repulsion of the two electrons, and then turning it on again very slowly. When you turn it off, the two blobs come together (it's the product of simple sine waves); then, when you allow the electrons to repel just a little bit, this blob can only deform in one way: it has to stretch towards the two corners just like the symmetric case. So that's what the solution has to look like. And in fact the energy of the anti-symmetric case turns out to be quite a bit higher.

I have to say it was a shock for me to contemplate that the shape of the solution might depend on how strong the repulsion was between the electrons. This really wasn't what I expected. And then it occured to me that if the solution depended on the strength of the repulsion, then it also depended on the size of the box. For real electrons, whose strength is constant, you get different waveshapes as you vary the size of the containment.

This is amazing. For the case of weak repulsion, the solution approaches the simple product function of two sine waves. In other words, both electrons fill the lowest energy state with little regard for what the other one is doing. Obviously for real electrons this must correspond to the case of the very large box.

But it doesn't! It turns out that the case of weak repulsion corresponds to the very tiny box. The repulsion only gets to be important as you make the box bigger and bigger. And a funny thing happens in the limit of the very big box: the symmetric and anti-symmetric solutions converge to the same energy! (In technical terms they become "degenerate".)

Tuesday, March 23, 2010

I promised to catch up on some topics I skipped over, so here goes. This problem arose when I was floundering on the helium atom; no, not the helium atom exactly, but the case of two hydrogen atoms far apart. I couldn't figure out why I kept getting "mini-helium" as a solution. So I backtracked to the simpler case of two potential wells. I thought that would help.

Then I realized I didn't know how to do even ONE potential well if it had a second electron in it. So I started playing around.

First I said, what if the two electrons make some minimal effort to stat away from each other. You know that the ground state is a sine wave, so I added a little “second harmonic” to each one. Something like so:

<!--[if !vml]--><!--[endif]-->

Now you see electron A is crowding towards the left, and electron B towards the right. So far so good. But the thing about two electrons, is they really want you to analyze them in terms of what they call the six-dimensional wave function in the vector coordinates of both electrons. At least for 3-d space, you get a six-dimensional wave function. For the one-dimensional well, you just get a two-dimensional plot, where the two axes are the coordinate for each electron. It looks something like this:

I've taken the wave function to be a direct product of A and B, which is the simplest thing you can do. It means other than the general crowding over, there is no particular detailed correlation between A and B. You can see from the graph that the wave function density is the highest when A is towards the left and B is towards the right.

The next step is a little interesting. In quantum mechanics you're not allowed to have a wave function that distinguishes in any meaningful way between two electrons. If you have A here and B there, you have to say that it's just as likely that B is here and A is there. So the function I've drawn is technically illegal. There is a way to make it right however, and it's called symmetrization. You just reverse A and B to get a new function. (It will look the same except the blob will be in the lower left corner instead of the upper right. Then you just add the two functions together. That way, the roles of A and B are automatically interchangeable:

And that's it: that's the hypothetical wave function for the two-electron well, chosen so that the electrons make some effort to stay away from each other, and symmetrized to comply with the requirement that the electrons are fundamentally indistinguishable.

It turns out there is another different way to symmetrize our wave function, but I'm going to have to wait til my next post to talk about that...

Thursday, March 18, 2010

It's hard for me to understand but I've done an awful lot of physics since I started this blog. Two blogs ago I said something about having some more stuff to post, and the next thing I posted was Quantum Siphoning. That's not what I was planning. It just came up all of a sudden, and it might actually be the solution to the long standing measurement problem, the so-called "collapse of the wave function". So it had to be posted.

But meanwhile a couple of cool problems got left aside. They were the fallout from my flirtation with mini-helium. One problem was the case of two electrons in a potential well. It has a surprising outcome. The other problem is the iso-electronic series of helium. It's also a two-electron problem, and the solution to the simple potential well pointed me in the right direction for the more complicated case. I actually wrote up the isoelectronic series and submitted it to a journal. But I'm still going to post it.

The one other very recent calculation that I haven't put up yet is my thermodynamic argument on how the reaction AgBr ---> Ag + 1/2 Br2 becomes spontaneous in the conditions of photographic exposure. That result led to my discovery of Quantum Siphoning, so I really should post the calculation. I'm getting around to it...

Damn, there's a lot of stuff I need to post. This is a calculation I did last year. The writeup it's pretty much barebones, but it's a very cool calculation.

EDIT: Right now (Jan 19 2014) this post is getting more hits than usual because of an upsurge of interest in the Ramanujan series. So I thought I'd revisit the whole thing and see if I couldn't explain it any better. You'll find my updated version here.)

Once in a long while I actually calculate something. I think I figured out how to do the Casimir effect. And I used that series 1 + 2 - 3 + 4....

First the series: it's one of those bizarre and counterintuitive results credited to the tragic Indian prodigy Ramanujan. For reasons that no one seems to know, he could intuitively recognize that certain alternating series which would appear to anyone else to be divergent actually had a finite sum. His conjectures were tested and almost always found to be correct.

I don't know how the mathematicians do it, but I can also make these series add up using my own little tricks. For example, the alternating arithmetic progression written above: you put it in an Excel spredsheet and then spread an envelope over it, a very gentle Gaussian. It quickly adds up to 0.25 which is the right answer. You can put the same Gaussian over the alternating squares, and it adds up to zero which is also correct. Of course you need a wider envelope to swallow up the much bigger terms.

The bizarre thing is how this weird mathematical series actually plays a role in modern physics. It can be used to calculate the Casimir Effect. Here's how the calculation goes.

We simplify things a little by putting everything in a one-dimensional box. So the energy modes are 1,2,3, etc. We can choose our units so that the pressure is numerically equal to the energy.

Now make the box twice small. The energy modes are 2,4,6...etc. But the PRESSURE is energy per unit volume (length, since it's one-dimensional)...so the pressures are (get this:) 4,8,12...

What's the difference in pressure between the big box and the small box?

1 + 2 + 3 .... - (4 + 8 + 12...)

which is...1 - 2 + 3 - 4 + 5 .... = 0.25!

If you take the pressure of the lowest mode in the smaller box to be 4, you see that the pressure difference between the small box and the big box is one sixteenth of the lowest-mode pressure. You can then take a chain of boxes, each one double the next one, to get the pressure relative to infinity: 1/16 + 1/64 + 1/256...= 1/12

And that's the calculatation. It's not really right for the three dimensional case, but I think if you apply it to a small cube and take the pressure defect to be 1/12 of the lowest mode pressure, you get something close to the ballpark for the parallel plate if you fill the gap with those small cubes. Or something like that. It's certainly true that the actual Casimir effect pressure is a definite fraction of the lowest mode pressure.

Wednesday, March 17, 2010

“The collapse of the wave function” has been with us for eighty-some years (almost a hundred if we include its immediate precursor, the “quantum leap”). It has to be considered one of the great philosophical bugaboos (?) of quantum mechanics and modern physics in general. But what exactly is it? If we could possibly agree on the definition of wave function collapse, I imagine it would have to be something close to the following: that in a physical process, governed by laws of nature in the form of differential equations, where the state of a system (the wave function) evolves continously through time, there comes a moment when something happens to the wave function at one place, and simultaneously, the wave function everywhere else simply ceases to exist. It’s a paradox.

Perhaps there are people who would find fault with the above characterisation of wave function collapse. No matter; there are surely countless examples which we could draw on to demonstrate the phenomenon. Unfortunately, I have not been able to find anywhere a published list of the 10 or 100 most compelling examples of wave function collapse. Quite the contrary; the Wikipedia article for example contains not a single concrete instance (although it does link to an article on Schroedingers Cat). So I have no choice but to go out on a limb and choose my own example.

I choose to focus on the specks of silver which appear on a photographic plate when it is exposed to the weak light of a distant star. Few would dispute that at least in the mass culture, this example would be near the top of almost anybody’s list. But in precisely what manner does it so convincingly epitomize the concept? We will find that question is not so easy to answer.

Superficially, the argument probably goes something like this. The reduction of metallic silver from silver bromide takes a certain amount of energy. This energy can only come the light of the distant star. We can consider light in the form of electromagnetic waves, and easily calculate its power density at the location of the silver atom. It is easy to show that this density, calculated over the cross section of the atom, is far too low to account for the chemical reduction of silver in any realistic time frame. Ipso facto, collapse of the wave function.

This argument is shaky on the grounds that it uses classical electromagnetic light as the driving force. Classical wave functions don’t collapse; only quantum mechanical ones do. In quantum mechanics, light is made of point particles called photons. There is no philosophical problem with a photon striking a silver atom and driving the conversion; it’s just a matter of probability. No need to collapse a wave function.

The resolution is of course to combine the two arguments. Particle or not, the photon is still governed by a quantum mechanical wave function which is essentially the same as the classical wave function. Until a speck of silver appears on the plate, the wave function of the photon is spread out everywhere. Then, at the moment the speck appears, the wave function of the photon vanishes everywhere. It has to vanish everywhere because all of its energy was required at that one point in space in order to drive the conversion, of which the visible evidence is the residual fleck of silver.

----------------------------

But there is another way in which one can challenge the idea of wave function collapse in the case of the photographic plate. Is it really true that the reduction reduction of silver from silver bromide requires the full measure of energy from a single photon? At first glance, the question seems almost absurd in its naivety. The enthalpy of the chemical reaction

AgBr ------------> Ag + ½ Br2

is 99 kJ/mol, or nearly one electron volt per atom of silver, and it is in the positive direction, which means it requires the input of energy.

However, it turns out that the detailed physical chemistry of the process is somewhat more complicated than the simple one-line reaction written above, and even today it is probably not fully understood to the last detail. It is said that the role of crystal defects and trace impurities is critical in the efficient functioning of photographic film.

Furthermore, in calculating the free energy of the reaction from the enthalpy above, we must not forget to include the effect of concentration. A silver bromide crystal containing trillions of atoms may be developable after exposure even if it contains a mere handful of reduced silver atoms; and there is a fascinating thermodynamic argument based the on the entropy of mixing which shows that at such concentrations, the point of equilibrium shifts so far to the right as to make it at least arguably plausible (plausibly arguable?) that there may be locations within the crystal where the reduction of silver is in fact thermodynamically favored.

All this must surely seem highly speculative. But it is upon this faint hope which the argument to follow hinges. I am suggesting the possibility that the appearance of the silver fleck represents the transition of the crystal from a less stable to a more stable state energetically. And therefore it did not require the full energy of the photon to procede.

Even this radical assumption does not solve the problem of “collapse of the wave function”! Let us see why not.

The problem has to do with the energy barrier associated with the transition. In accordance with our proposed suspension of disbelief, we are going to suppose that the exposed AgBr crystal with its reduced silver atom is in a lower energy state than the undisturbed crystal. If this were the whole story, we would not need the energy of a photon to drive the transition. But the problem is in getting from A to B. According to our best understanding, the incident photon liberates a valence band electron by promoting it to the conduction band. I would like to say definitively how much energy this takes, but I have not been able to find this data. It is certainly a known fact that photographic film can be safely exposed to the deep red light of a dark room, but I haven’t found the actual cutoff frequency. Let’s assume it is 2 eV for the sake of argument.

The problem is that once we assume the photon is absorbed in promoting the electron to the valence band, we have implicitly assumed the collapse of the photon’s wave function. And that’s just what I’m trying to avoid.

The goal is to follow the process through the time evolution of the Schroedinger wave function to see if we can bring it to completion without at any stage invoking the collapse of the wave function. To facilitate this analysis, I have simplified the system to what I hope will be most easily manageable while retaining the essential features of the actual process.

What I have analyzed is the case of two potential wells inside a box. One well (A) is at a slightly higher potential than the other (B), and it is at (A) that the electron is trapped. Both wells are located inside a larger box (C) so the problem is confined to a finite volume.

So far, so good. But in what follows, it is a little tedious to have to remember which is A, which is B, and which is C. So I’m going to rename them to make things a little easier to follow. Remember, these are just names, and they aren’t to be taken as meaning any more than that.

We will refer to the well at A as “the silver halide”. The well at B, at a slightly lower potential , will be called “the silver atom”. And the big box, C, will be called “the conduction band”. Remember, despite what we call them, they are just two potential wells inside a box, and the similarities to any real chemical processes are, if not purely coincidence, at least incomplete. Oh, one more thing: the potential required to promote an electron from the “silver halide” to the “conduction band” (you see how this is going to work?) will be called “the band gap”.

We will assume that for one reason or another, the electron cannot get from the silver halide to the silver atom by tunneling. In our model we can arrange for this by having the two wells sufficiently far apart. We will further stipulate that the quantum of energy required to get into the conduction band is 2 eV.

What I proposed was that we consider what happens if instead of absorbing all the energy from a discrete photon, silver halide absorbs from a continuous electromagnetic field, only a fraction of that quantum of energy before the field disappears. Then the electron is then left in a superposition of states; say, 90% silver halide and 10% conduction band.. Since the conduction band is also coupled to the silver atom, the wave function may evolve further so that the electron finds itself in a superposition of 90% silver halide, 10% conduction band and 1% silver atom. (Due to rounding, percentages may not add to exactly 100.) And now the question: where does the system go from here? In particular, is it possible for the system to end up with the electron entirely in the silver atom?

Using the Born rule, standard QM would seem to tell us that there is a 90% probability of finding it in the silver halide, a 10% chance of finding in the conduction band, and a 1% chance of finding it in the silver atom. But that is not especially helpful. For one thing, as long as the conduction band is partially filled, there will be radiating energy which we are going to assume can escape from the box. And as the radiant energy is lost from the system, the conduction band will be depleted to the benefit of both the silver halide and the silver. We might then expect the system to stabilize in the proportions of 90/10 silver halide to silver. The exact percentages don’t matter; the point is, there is a probability of finding the electron in the silver atom.

But the Born rule which we used to get those probabilities essentially demands the collapse of the wave function, and that’s just the thing we’re trying to avoid. We avoided it at the point the photon was absorbed by saying we wouldn’t take a whole photon, we’d just take a portion of a continuous e-m wave. It defeats our whole purpose to now invoke the Born Rule to force it into the silver atom. We return to the question: can the electron get from the mixed state of silver/silver halide/condutcion band to a state where it is entirely in the silver atom, and can it get there by means of a process which evolves naturally in space and time?!!

There is enough energy available to drive the process; that’s not the problem. When the conduction band couples to the silver atom, it is true that the probability drains toward the silver which is what we want. And energy is released which we would like to use in order to replenish the silver halide, to keep the process moving. Very much like an ordinary siphon. The probelm is that the energy at the silver atom is released in the form of radiating electromagnetic energy, and it was not clear to me how could I recapture it with good efficiency at the silver halide, especially if the two are relatively far apart.

I though about this for a long time and couldn’t get around it. Then the answer came to me. You can’t do it. You can’t recapture the energy lost radiated away from the silver atom, receive it at the silver halide, and pump it back up to the conduction band.

But you can do it if you have millions and millions of silver halide sites! We will call such a collection of potential wells (remember, that’s all they are) a “crystal”. In the very middle of the crystal is one special well, the “silver atom”, just a little deeper than all the other wells, and it is empty.

Now pass a wave of light through the “crystal”. Not a very strong wave, but at a frequency sufficient to couple the silver halides to the conduction band. When the wave is gone, the conduction band is excited to the extent of 10%, and all the silver halides ground state wave functions are filled to the extent of only 99.99999%.

Notice carefully that according to my picture, you haven’t captured enough energy to put a whole electron into the conduction band. Just 10% of one. And yet now I’m going to show how the system will evolve through time so that the electron ends up in the silver atom. The process is what I call quantum siphoning.

It’s not hard to get a little bit of wave function excitation happening in the silver atom. There’s plenty available in the conduction band to excite it to the level of, say 1%. But as it starts to fills up, the conduction band is emptying out, and it is not being replenished. The energy which was available is simply being radiated away from the site of the silver atom.

But the silver atom is surrounded by millions of receiving antennas which are tuned to its exact frequency. These are none other than all the surrounding silver halides. As the electromagnetic wave radiates outwards, it can’t help but excite the millions of potential wells in its path. And as it does, it leaves a little of its energy behind at each one, getting weaker and weaker as it goes. At the same time a wee liitle portion of all the electron wave functions in those millions of silver halides is is promoted to the conduction band. In fact, the penetration depth of the em wave emanating from the silver atom decays exponentially, until essentially one hundred percent of it is absorbed. And every bit of that absorbed energy goes towards replenishing the conduction band. There is literally nowhere else for it to go. It is in fact the quantum mechanical version of a siphon. The replenished conduction band can now continue to refill the silver atom ground state until it’s completely full. Full of exactly one electron.

Let us recap: A weak, diffuse electromagnetic wave passed through the crystal. A small portion of a quantum of energy was absorbed in the passage, and the effect of this absorption was recorded by the appearance of an electron in the silver atom. Every step of the process occured according to the ordinary time evolution laws of the Schroedinger and Maxwell equations, and at no point was it necessary for a wave to “collapse to an eigenfunction of the state being measured.”

And finally, not to argue the point to strongly, although it is just a though experiment with potential wells scattered in a big box, it does seem to shares some of the characteristic features of the traditional photographic plate, with its flecks of silver appearing one by one under the influence of very weak starlight. Is light made of photons? Does the wave function collapse? Or can we not perhaps explain it all by wave-on-wave interactions according to the well-known laws of Schroedinger and Maxwell?

Sunday, March 14, 2010

It's been a while, but I finally got to the bottom of my problem with the helium atoms. You remember that I had two isolated protons and I tried to solve the Schroedinger equation by sharing two electrons between them so each atom looked like a miniature helium. Now I know what I did wrong.

You can account for the energy of the system by adding up five terms. They are:

(1) the kinetic energy of electron A (2) the kinetic energy of electron B (3) the potential energy of electron A (4) the potential energy of electron B (5) the repulsion energy of electron A versus electron B

If you have a solution to the Shroedinger equation, and you make a new wave function where all these terms are exact multiples of your old solution, then the new wave function will also be a solution. That's what I was trying to do.

I took the helium atom solution and spread it out in space so it was twice as wide. Then I cloned it and put one replica at proton A and one replica at proton B. Looking at the five components of system energy, it appeared to me that each one was exactly one quarter of the original, giving me a valid solution. That was my mistake.

The kinetic energy of electron A is indeed one quarter of the original, and so is the kinetic energy of electron B. It works because the del-squared operator automatically gives you one-quarter the result when you double the linear dimension.

The potential energy of electron A is also one quarter of the original, as is the potential energy of electron B. It works because at each atom you have one-eighth the energy: half the nuclear charge, half the electron charge, and twice the distance. At first glance you might think there ought to be extra terms in the potential energy on account of the attraction of proton A for electron B and vice versa, but I can reduce these terms arbitrarily close to zero by putting the atoms far apart. No, the potential energy works out OK. It is the repulsion energy which is messed up.

The repuslion energy of the two electrons appears at first glance to work out exactly the same as the potential energy. At each atom you have half an electron repelling half an electron at twice the distance: one-eighth the energy. Double it for the second atom and you are back to one quarter, so everything seems proportional. But it isn't.

I am not a fan of the probability density interpretation of the wave function but in this instance I don't have a better explanation. The interpretation that works is not that you have half an electron repelling half an electron. It is that you have a 50% probability of a whole electronrepelling a whole electron. This gives you twice the energy as what I calculated, so this term goes out of whack with the other four terms.

It has to work this way because otherwise, you could apply this technique in the opposite direction and solve the doubly ionized beryllium atom (Be++) as a squeezed-down replica of the helium atom. All the energy levels would be exactly four times as big. In fact you do just this when going from the hydrogen atom to the He+ ion. It works in that case because with only oneelectron there is no repulsion term. The isoelectronic series of hydrogen consists scaled copies of the identical wave function. But the isoelectronic series of helium doesn't work that way.

So I can't create mini-helium by sharing two electrons between two isolated protons. But that doesn't mean my problem doesn't have a solution. It just means that the wave function I chose does not minimize the energy of the system. There is a solution, and it is in the shape of the hydrogen negative ion, a little-known form of hydrogen with an extra electron. It seems that there is just enough attraction between a hydrogen atom and a free electron to make this a stable species.

So it means you can take the wave function of H- and clone it so each proton gets a copy. Then you share the two electrons between the two protons. It looks strange but it's a solution of the Schroedinger equation. Each proton has two "half-electrons" bound to it.

There's a more conventional solution where both electrons go to one proton and the other proton sits there all alone. That's the familiar solution. Is there a relationship between the two solutions?

Yes, and it's called symmetrization. It's something you actually do all the time in quantum mechanics. You observe that there is nothing special about A or B, so any solution which distinguishes them must have a counterpart where the roles are reversed. You take these two complementary solutions and make a new solution by adding them together. That's called symmetrization and it gives you my distributed mini-ions. Using the hydrogen negative ion of course rather than helium.

There is another side to symmetrization: if you can take the sum of two wave functions you can also take the difference. It's called, not without some logic, "anti-symmetrization". In quantum mechanics you can always do that with states: instead of working with states A and B, you work with the sums and differences: A+B (symmetric) and A-B (antisymmetric). What if you're really interested in what happened to state A? Easy: just apply the same trick to your new states. You can see that if you add the symmetric and antisymmetric states together, it just returns you to state A. The difference gives you back state B.

When I started this blog a month ago I said I was fed up with physics because I couldn't solve a single problem with two electrons in it. The funny thing is maybe I have now. Not just this problem but a couple more, which I'm going to talk about in my next post.